Proof of n - \lceil \dfrac{n}{2} \rceil = \lfloor \dfrac{n}{2} \rfloor (Latex test - MathJax)

The key here is to divide the cases, it is not really difficult at all

To proof : n - \lceil \dfrac{n}{2} \rceil  = \lfloor \dfrac{n}{2} \rfloor

Case 1.  n = 2k + 1

n - \lceil \dfrac{n}{2} \rceil  = 2k + 1 - ( k+1) = \lfloor k + \dfrac{1}{2} \rfloor = k

Both the LHS and RHS are k therefore, the equality stands true.

Case 2.  n = 2k

n - \lceil \dfrac{n}{2} \rceil  = 2k - (k) = \lfloor \dfrac{2k}{2} \rfloor  = k

Both the LHS and RHS are k therefore, the equality stands true.

Thus, we have shown  n - \lceil \dfrac{n}{2} \rceil  = \lfloor \dfrac{n}{2} \rfloor

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